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Home > Maths II > Numbers and money > Compound interest and appreciation/depreciation

Maths II

Compound interest and appreciation/depreciation

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Cumulative increase and decrease

Simple interest

With simple interest the amount of money borrowed remains fixed.

For example £400 is borrowed for 3 years at an interest rate of 5% pa (pa means per annum, or each year).

Interest for one year = 5% of £400

= ({5 \over 100}) \times 400

=\pounds 20

Interest for 3 years = \pounds 20 \times 3 = \pounds 60

You can write this in a formula.

P \times R \times T

  • P (principal) is the amount borrowed.
  • R is the rate of interest per year.
  • T is the time in years.

Compound Interest

Here the interest is added to the principal at the end of each year. So the next year the interest is worked out on a larger amount of money than what was originally borrowed.

This means paying interest on the interest of previous years (unlike simple interest, where you only pay interest on the original amount).

This is how it is calculated:

£400 is borrowed for 3 years at 5% compound interest.

A photo of someone filling out a mortgage application form.

Principal at the start = £400

Interest in the 1st year = {5 \over 100} \times 400 = \pounds 20

Principal after 1 year = £420

Interest in the 2nd year = {5 \over 100} \times 420 = \pounds 21

Principal after 2 years = £441

Interest in the 3rd year = {5 \over 100} \times 441 = \pounds 22.05

Principal after 3 years = £463.05

The total interest charged under compound interest will be £63.05.

This is different to the simple interest worked out above.

Four types of question

In percentage questions, read the question carefully and decide what you are being asked to do. You may need to:

  • Find a given percentage of an amount.
  • Work out a percentage when given 2 amounts.
  • Work backwards from a percentage increase or decrease (reverse percentages).
  • Find a cumulative change.
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